Question

Solve $\frac{dy}{dx}+2y\tan x=\sin x.$ If at $x=\frac{\pi }{3}$, $y=0$ then

• A. $y=\cos x+2{\cos }^{2}x.$
• B. $y=-\cos x-2{\cos }^{2}x.$
• C. $y=\cos x-2{\cos }^{2}x.$
• D. None of these.

Answer 1

The correct option is C $y=\cos x-2{\cos }^{2}x.$

$\frac{dy}{dx}+2y\tan x=\sin x$ ...(1)

Here $\int Pdx=\int 2\tan xdx=2\log \sec x=\log {\sec }^{2}x$

$\therefore I.F.=e^{\log {\sec }^{2}x}={\sec }^{2}x$

Multiplying (1) by $I.F.$, we get

${\sec }^{2}x\frac{dy}{dx}+2y\tan x{\sec }^{2}x=\sin x.{\sec }^{2}x$

Integrating both side, we get

$y{\sec }^{2}x=\int \sin x{\sec }^{2}xdx+c=\int \tan x\sec xdx+c=\sec x+c$

$\Rightarrow y=\cos x+c{\cos }^{2}x$

Given at $x=\frac{\pi }{3}$, $y=0$

$\therefore 0=\frac{1}{2}+c.\frac{1}{4}\Rightarrow c=-2$

Hence $\therefore y=\cos x-2{\cos }^{2}x$